To 15.50 mL of an unknown NaOH solution is added 17.00 mL of 0.460 M H2SO4 solution. The resulting mixture is back titrated with 2.20 mL of 0.1100 M NaOH to a phenolphthalein end-point. What was the molarity of the NaOH in the unknown?
H2SO4 + 2NaOH ==> Na2SO4 + 2H2O
NOTE: THESE CALCULATIONS ARE APPROXIMATE TO SHOW YOU HOW TO WORK THE PROBLEM BUT THEY ARE NOT AS ACCURATE AS THEY NEED TO BE SO YOU WILL NEED TO RECALCULATE EACH STEP.
Total millimoles H2SO4 = mL x M = 17.70 x 0.460 = 8
mmoles NaOH for back titration = 2.20 x 0.1100 = 0.2
0.2 mmoles NaOH back titration = 0.1 mmoles H2SO4
Subtract 8-0.1 = 7.9 mmols H2SO4 used for the titration of the unknown.
Since the H2SO4/unknown NaOH is a 2:1 ratio, then 7.9/2 = about 4 mmoles NaOH in the unknown. Then M = millimiles/mL = about 4/15.50 = ?
I have ignored significant figures, obviously, but you have 3 in the volumes so you should have 3 throughout unless those 2.20 and 0.460 values actually are 2.200 and 0.4600.
Post your work if you get stuck.